Question

A ball thrown by one player reaches the other in `2 s`. The maximum height attained by the ball above the point of projection will be about.
A. `2.5 m`
B. `5m`
C. `7.5 m`
D. `10m`

Answer 1

Correct Answer - B
Since, the ball reached from one player to another in 2s, so the time period of the flight, `T = 2s`
`rArr (2u sin theta)/(g) = 2s`
Here, u is the initial velocity and `theta` is the angle of projection,
`rArr u sin theta = g` .....(i)
Now, er know that the maximum height of the projection
`H = (u^(2)sin^(2)theta)/(2g)` or `H = ((u sin theta)^(2))/(2g)`
On putting the value of `u sin theta` from Eq. (i), we have
`H = (g^(2))/(2g) = (g)/(2)` or `H = (g)/(2) = (10)/(2)m` or `H = 5m`